Ask question

Ask question

All categories

2 years ago

14

How many valence electrons does rubidiumn have?

1 answer:

vlada-n [284]2 years ago

7 0

Answer:

1

Explanation:

I believe it really only has 1 valence electrons

You might be interested in

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago

In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2

vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory was modified and extended by Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

$\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}$

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

$H_2O$ accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

8 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

Urgent help please!!

sveta [45]

Answer:

1. 2.1 moles of Mg

2. 0.72 mole of Mg(OH)2

Explanation:

1. We'll begin by writing the balanced equation for the reaction. This is given below:

3Mg + 2AlBr3 —> 3MgBr2 + 2Al

From the balanced equation above, 3 moles of Mg reacted to produce 2 moles of Al.

Therefore, Xmol of Mg will react to produce 1.4 moles of Al i.e

Xmol of Mg = (3 x 1.4)/2

Xmol of Mg = 2.1 moles.

Therefore, 2.1 moles of Mg is required to 1.4 moles of Al.

2. We'll begin by calculating the number of mole in 26g of water, H2O.

This is illustrated below:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O = 26g

Number of mole of H2O =?

Mole = Mass /Molar Mass

Number of mole of H2O = 26/18

Number of mole of H2O = 1.44 moles

Next, we shall write the balanced equation for the reaction. This is given below:

2HNO3 + Mg(OH)2 —> Mg(NO3)2 + 2H2O

Finally, we can obtain the number of mole of Mg(OH)2 used in the reaction as follow:

From the balanced equation above,

1 mole of Mg(OH)2 reacted to produce 2 mole of H2O.

Therefore, Xmol of Mg(OH)2 will react to produce 1.44 moles of H2O i.e

Xmol of Mg(OH)2 = (1 x 1.44)/2

Xmol of Mg(OH)2 = 0.72 mole.

Therefore, 0.72 mole of Mg(OH)2 was used in the reaction.

3 0

3 years ago

Lithium peroxide chemical formula ?

Andreyy89

<span>Li2O2 is the formula

</span>

7 0

3 years ago

Read 2 more answers